College Algebra
posted by Nancy on .
A coin jar contains nickels, dimes and quarters. There are 46 coins in all. There are 11 more nickels than quarters. The value of the dimes is $1.00 less than the value of the quarters. How many coins of each type are in the jar?

Solve the system:
n+d+q=46
n=q+11
10d=25q100
From the 3rd >d=2.5q10
The 1st: (q+11)+(2.5q10)+q=46
q=10
n=21
d=15 
solving using the elimination method and identify the system as consistent, inconsistent or dependent.
3x4y=8
6x2y=10 
Suppose that Maria has 120 coins consisting of pennies, nickels, and dimes. The number of nickels she has is 14 less than twice the number of pennies; the number of dimes she has is 22 less than three times the number of pennies. How many coins of each kind does she have?

p+n+d=120
n=2p14
d=3p22
p+(2p14)+(3p22)=120
6p36=120
6p=156
p=26
n=2(26)14
n=5214
n=38
d=3(26)22
d=7822
d=56 
P(1, 2); Q(8, 26)

5+6